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A uniform metal wire carries a current of $2 \mathrm{~A}$ when an ideal cell of $3.4 \mathrm{~V}$ is connected across it. The wire has mass $8.92 \times 10^{-3} \mathrm{~kg}$, density $8.92 \times 10^3 \mathrm{kgm}^{-3}$ and resistivity $1.7 \times 10^{-8} \Omega \mathrm{m}$. Then the length of the wire is
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The correct answer is:
$10 \mathrm{~m}$
Current, $\mathrm{I}=2 \mathrm{~A}$
Potential, $\Delta V=3.4 \mathrm{v}$
Using Ohm's law
$\begin{aligned}
& \mathrm{R}=\frac{\Delta \mathrm{V}}{\mathrm{I}}=\frac{3.4}{2}=1.7 \Omega \\
& \text { Volume }=\frac{\text { Mass }}{\text { Density }} \\
& =\frac{8.92 \times 10^{-3}}{8.92 \times 10^3}=10^{-6} \mathrm{~m}^3 \\
& \mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{VA}} \times \frac{\mathrm{L}}{\mathrm{L}}=\frac{\rho \mathrm{L}^2}{\mathrm{~V}} \\
& \mathrm{~L}^2=\frac{\mathrm{RV}}{\rho}=\frac{1.7 \times 10^{-6}}{1.7 \times 10^{-8}}
\end{aligned}$
Length of the wire, $\mathrm{L}=10 \mathrm{~m}$.
Potential, $\Delta V=3.4 \mathrm{v}$
Using Ohm's law
$\begin{aligned}
& \mathrm{R}=\frac{\Delta \mathrm{V}}{\mathrm{I}}=\frac{3.4}{2}=1.7 \Omega \\
& \text { Volume }=\frac{\text { Mass }}{\text { Density }} \\
& =\frac{8.92 \times 10^{-3}}{8.92 \times 10^3}=10^{-6} \mathrm{~m}^3 \\
& \mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{VA}} \times \frac{\mathrm{L}}{\mathrm{L}}=\frac{\rho \mathrm{L}^2}{\mathrm{~V}} \\
& \mathrm{~L}^2=\frac{\mathrm{RV}}{\rho}=\frac{1.7 \times 10^{-6}}{1.7 \times 10^{-8}}
\end{aligned}$
Length of the wire, $\mathrm{L}=10 \mathrm{~m}$.
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